The phase difference between voltage and current in series L-C circuit is

Options

A. `0^(@)`
B. `pi`
C. `(pi)/(2)`
D. `(pi)/(3)`

Answer

To find the phase difference between voltage and current in a series L-C circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Components**: In a series L-C circuit, we have an inductor (L) and a capacitor (C) connected in series. The resistance (R) in this case is considered to be zero.

2. **Phase Angle Formula**: The phase angle (φ) in an L-C-R circuit is given by the formula:
\[ \phi = \tan^{-1} \left( \frac{X_L - X_C}{R} \right) \]
where \(X_L\) is the inductive reactance and \(X_C\) is the capacitive reactance.

3. **Set Resistance to Zero**: Since we are dealing with a pure L-C circuit, we set \(R = 0\). Thus, the formula becomes:
\[ \phi = \tan^{-1} \left( \frac{X_L - X_C}{0} \right) \]

4. **Evaluate the Expression**: When \(R = 0\), the expression becomes:
\[ \phi = \tan^{-1} \left( \text{positive or negative value} \right) \text{ divided by } 0 \]
This leads to an undefined situation, which indicates that the phase angle approaches infinity.

5. **Interpret the Result**: When the phase angle approaches infinity, it implies that the current lags the voltage by \(90^\circ\) or \(\frac{\pi}{2}\) radians. Therefore, we can conclude:
\[ \phi = \frac{\pi}{2} \text{ or } 90^\circ \]

6. **Final Answer**: The phase difference between voltage and current in a series L-C circuit is:
\[ \phi = \frac{\pi}{2} \text{ radians or } 90^\circ \]